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p-Block Elements Test - 7

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p-Block Elements Test - 7
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Weekly Quiz Competition
  • Question 1
    4 / -1

     

    Which of the following is/are correct statement(s) regarding noble gases?

    i. He2 molecule does not exist

    ii.  value for noble gases is 1.4
    iii. Heaviest noble gas present in air is xenon.
    iv. 6th period starts with caesium and ends with radon.

     

    Solution

     

     value for noble gases is 1.66

     

  • Question 2
    4 / -1

    Which statement is false?

    Solution

    These three gases were discovered by Travers and Ramsay

  • Question 3
    4 / -1

    The ionisation potential of hydrogen atom is 13.6 eV. Then ionisation potential of He+ ion is

    Solution

     For hydrogen like atom

  • Question 4
    4 / -1

    Which of the following is/are correct among the following?

    i. XeF2 = sp3d, linear
    ii. XeF4 = sp3, square planar
    iii. XeO4 = sp3, tetrahedral
    iv. XeF6 = sp3d2, octahedral

    Solution

     Hybridisation of the compounds are

  • Question 5
    4 / -1

    The oxidation number of oxygen in O2PtF6 is

    Solution

    The structure of O2PtF6 is  O+2 [PtF6]-.
    Oxidation tate of O2 = 2x = +1
    x = + 1/2

  • Question 6
    4 / -1

    The structure of  ion is

    Solution

  • Question 7
    4 / -1

    The number of (pπ-dπ) π-bonds present in XeO3 and XeO4 respectively are 

    Solution

  • Question 8
    4 / -1

    XeF6 on reaction with CsF produce

    Solution

    The correct answer is option D
    CsF + XeF6 → Cs[XeF7]
     

  • Question 9
    4 / -1

    Which of the following reaction occurs?

    Solution

    All the cases are possible reactions.

  • Question 10
    4 / -1

    The wrong statement among the following is

    Solution

    He is used in atomic reactor for cooling.

  • Question 11
    4 / -1

     

    Which oxide of nitrogen is obtained on heating ammonium nitrate at 250oC?

     

    Solution

     

     

  • Question 12
    4 / -1

     

    Which of the following oxides is anhydride of nitrous acid?

    Solution

     

     2HNO2 → N2​O3​ + H2O

  • Question 13
    4 / -1

     

    A translucent white waxy solid (A) reacts with excess of chlorine to give a yellowish white powder (B). (B) reacts with organic compounds containing -OH group converting them into chloro derivatives. (B) on hydrolysis gives (C) and is finally converted to phosphoric acid. (A), (B) and (C) are?

    Solution

     

    A translucent white waxy solid (A) reacts with excess of chlorine to give a yellowish white powder (B). Here B is PCl5​.PCl5​ reacts with organic compounds containing -OH group converting them into chloro derivative POCl3​..PCl5​ on hydrolysis gives (C).C is POCl3​ and is finally POCl3​ converted to phosphoric acid.

    So  (A), (B) and (C) are P4​,PCl5​,POCl​3.

     

  • Question 14
    4 / -1

     

    Which of the following statements is not correct about the structure of PCl5​?

    Solution

     

    The axial bond pairs suffer more repulsion as compared to equatorial bond pairs.

     

  • Question 15
    4 / -1

     

    Why are all P-F bonds in PF5 are not equivalent?

    Solution

    Structure of PF5

    PF5 has three axial bonds and two equitorial bonds and has sp3d hybridisation PCl5 has trigonal bipyramidal structure .

    Due to this two Cl atoms lie along axial line and other three Cl atoms lie along the equatorial plane.

    Hence Cl atoms which lie along axis have different bond length than that of Cl atoms lying on equatorial plane.

     

  • Question 16
    4 / -1

    Direction (Q. Nos. 16 and 17) This section contains a paragraph, each describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).

    Passage

    The noble gases have closed-shell electronic configuration and are monoatomic gases under normal conditions. The lower boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.
    Direct reaction of xenon with fluorine leads to the formation of a series of compounds with oxidation numbers +2, + 4, + 6 and + 8. XeF4 reacts violently with water to give XeO3. The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell.

    Q. 

    The noble gas compound without lone pair on xenon atom is

    Solution

  • Question 17
    4 / -1

    The noble gases have closed-shell electronic configuration and are monoatomic gases under normal conditions. The lower boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.
    Direct reaction of xenon with fluorine leads to the formation of a series of compounds with oxidation numbers +2, + 4, + 6 and + 8. XeF4 reacts violently with water to give XeO3. The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell.

    Q. 

    Which of the following property decreases from helium to xenon?

    Solution

    Rest of properties increases from He to Xe.

  • Question 18
    4 / -1

     

    Direction (Q. Nos. 18 and 19) This section is based on Statement I and Statement II. Select the correct answer from the codes given below.

    Q. 

    Statement I : Radon is produced upon α-decay of radium atom.

    Statement II : α-particles which are equivalent to the atom, reduces the atomic number by 2 and atomic mass by 4 upon decay of mother nucleus.

     

  • Question 19
    4 / -1

    Statement I : The structure of XeF4 is square pyramidal.

    Statement ll : The hybridisation of XeF4 is sp3d2.

    Solution

    Structure of XeF4 is square planar.

  • Question 20
    4 / -1

    Direction (Q. No. 20) Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d) out of which one is correct.

    Q. 

    Match the Column I with Column II and mark the correct option from the codes given below.

    Solution

  • Question 21
    4 / -1

     

    Direction (Q. Nos. 21-25) This section contains 5 questions. When worked out will result in an integer from 0 to 9 (both inclusive).

    Q. 

    Number of lone pairs on the central atom in XeOF4 are

     

    Solution

     


     

  • Question 22
    4 / -1

     

    Among XeF2, XeF4, XeF6, XeO3, XeO4, XeO2, XeOF4 the number of compounds that have non-zero dipole moment.

     

    Solution

    Correct Answer :- 4

    Explanation : XeF4 is square planar and XeF2 is linear in shape. The molecules are symmetrical and so the vector sum of the individual dipole moment is zero.

    XeO3 is planar in shape. It has 3 dative bonds(coordinate bonds) in one single plane. Taking into consideration that dipole moment is a vector, the 3 dipole moments of 3 bonds being in same plane and of equal magnitude results in a sum of zero. Therefore the dipole moment of net molecule is 0.

     

  • Question 23
    4 / -1

     

    In XeF2, state the total number of electron pairs in its Lewis dot structure.

     

    Solution

     

     Electron dot structure of XeF2 is

     

  • Question 24
    4 / -1

     

     In argon, find the number of electrons with spin value + 1/2.

     

    Solution

     

    Electronic configuration of Ar is

    Here, 9 electrons each has + 1/2 and - 1/2 value of magnetic moment spin.

     

  • Question 25
    4 / -1

     

    Find the number of unpaired electrons in the fully excited xenon atom.

     

    Solution

     

    The electronic configuration of Xe is [Kr]4d105s25p6 depending on the energy provided 8 electrons (from 5s and 5p) can go into excited state is 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s, etc.

     

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